# Aerodynamic Drag

## Discussion

### pressure drag

The force on an object that resists its motion through a fluid is called drag. When the fluid is a gas like air, it is called aerodynamic drag (or air resistance). When the fluid is a liquid like water it is called hydrodynamic drag (but never "water resistance").

Fluids are characterized by their ability to flow. In somewhat technical language, a fluid is any material that can't resist a shear force for any appreciable length of time. This makes them hard to hold but easy to pour, stir, and spread. Fluids have no definite shape but take on the shape of their container. (We'll ignore surface tension for the time being. It's really only significant on the small scale — small like the size of a droplet.) Fluids are polite in a sense. They yield their space relatively easy to other material things; at least when compared to solids. A fluid will get out of your way if you ask it. A solid has to be told with destructive force.

Fluids may not be solid, but they are most certainly material. The essential property of being material (in the classical sense) is to have both mass and volume. Material things resist changes in their velocity (this is what it means to have mass) and no two material things may occupy the same space at the same time (this is what it means to have volume). The portion of the drag force that is due to the inertia of the fluid — the resistance that it has to being pushed aside — is called the pressure drag (or form drag or profile drag). This is usually what someone is referring to when they talk about drag.

Recall Bernoulli's equation for the pressure in a fluid…

*P*_{1} + ρ*gy*_{1} + ½ρ*v*_{1}^{2} = *P*_{2} + ρ*gy*_{2} + ½ρ*v*_{2}^{2}

The first term on each side of the equation is the part of the pressure that comes from outside the fluid. Typically, this refers to atmospheric pressure weighing down on the surface of a liquid (not relevant right now). The second term is the gravitational contribution to pressure. This is what causes buoyancy (also not relevant right now). The third term is the kinetic or dynamic contribution to pressure — the part related to flow (very relevant). This will help us understand the origin of pressure drag.

Start with the definition of pressure as force per area. Solve it for force .

P = |
F |
⇒ | F = PA |

A |

Replace the generic force symbol *F* with the more specific symbol *R* for drag. (You can also use *D* if you wish.) Drop in Bernoulli's equation for the pressure in a moving fluid…

F = PA = |
⎛ ⎝ |
1 | ρv^{2} |
⎞ ⎠ |
A |

2 |

Rearrange things a bit and here you go…

*R* = ½ρ*CAv*^{2}

Wait a minute. Where'd that extra symbol come from? Who put that *C* in there and why?

Let's run through all the symbols one at a time, explain their meaning and how they relate to pressure drag. In essence, let's take the equation apart and put it back together again.

- Drag increases with the
*density*of the fluid (ρ). More density means more mass, which means more inertia, which means more resistance to getting out of the way. The two quantities are directly proportional.*R*∝ ρ - Drag increases with
*area*(*A*). Exactly what we mean by this is subject to debate. To me, and in the context of this model, area is the cross sectional area projected in the direction of motion. (I would further simplify this by calling it the projected area.) Take the cross section of the object in the direction of its motion. This is the area of the tube of fluid that must be cast aside to let the object pass. This is the most logical thing to call the area, but not everyone agrees with me. To some, the word "area" refers to the area of contact between the object and the fluid. This also makes sense, but not in the context I've described above. Surface area is not important when one is dealing with pressure drag, but it is important when dealing with viscous drag — drag caused by layers of the fluid sticking to the object and to one another. More surface area means more of the object is in contact with the fluid, which means more drag. Viscous drag is just as real as pressure drag, but I don't want to deal with it right now.*R*∝*A* - Drag increases with
*speed*(*v*). I hope that this is self-evident. An object that is stationary with respect to the fluid will certainly not experience any drag force. Start moving and a resistive force will arise. Get moving faster and surely the resistive force will be greater. The hard part of this relationship lies in the detailed way speed affects drag. According to our very sensible model derived from Bernoulli's very sensible equation, drag should be proportional to the square of speed.*R*∝*v*^{2}

Which brings us to our last factor… - Drag is influenced by other factors including shape, texture, viscosity (which results in viscous drag or skin friction), compressibility, lift (which causes induced drag), boundary layer separation, and so on. These factors can be dealt with separately in a more complete theory of drag (how tedious in one sense, but how necessary in another) or they can be piled into one monolithic fudge factor (oh yes, please) called the coefficient of drag (
*C*).*R*∝*C*

Combining all these factors together yields a theoretically limited (but empirically very reasonable) equation. Here it is again…

*R* = ½ρ*CAv*^{2}

Simple, compact, wonderful. A nice equation to work with — or is it?

Well, yes and no.

- Yes, but it works only as long as the range of conditions examined is "small". That is, no large variations in speed, viscosity, or crazy angles of attack. The way around this is to reduce the coefficient of drag to a variable rather than a constant. (I can live with this.) Say that
*C*depends on some yet to be specified set of factors. It is totally acceptable to say that it varies with this that or the other quantity according to any set of rules determined by experiment. - No, since speed is squared. [Gasp!] Recall that speed is the derivative of distance with respect to time. Have you ever tried to solve a nonlinear differential equation? No? Well, welcome to hell. Wait, let me rephrase that — Welcome to Hell! [Ca-rack! Boom!] Ah ha ha ha ha haaaa! [Rumble] You fool! Just wait till you see what's in store for you when you try to solve the differential equations. The mathematics will consume you. [Ca-rack! Boom!] Ah ha ha ha ha haaaa! [Rumble].

Whew. What the hell was that all about? I might not know how to solve every kind of differential equation off the top of my head, but so what. I can always look for the solution in a book of standard mathematical tables or an on-line equivalent. You don't scare me demonic voice in my head.

C_{d} |
object or shape |
---|---|

2.1 | ideal rectangular box |

1.8~2.0 | eiffel tower |

1.3~1.5 | empire state building |

1.0~1.4 | skydiver |

1.0~1.3 | person standing |

0.9 | bicycle |

0.7~1.1 | formula one race car |

0.6 | bicycle with faring |

0.5 | ideal sphere |

0.7~0.9 | tractor-trailer, heavy truck |

0.6~0.7 | tractor-trailer with faring |

0.35~0.45 | suv, light truck |

0.25~0.35 | typical car |

0.15 | Aptera high-efficiency electric car |

0.15 | airplane wing, at stall |

0.05 | airplane wing, normal operation |

0.020~0.025 | airship, blimp, dirigible, zeppelin |

### other mathematical models

The pressure drag equation derived above is to me the most reasonable mathematical model of drag — especially aerodynamic drag. But as the demonic voice in my head said, it isn't always the easiest one to work with — especially for those just learning calculus (differential equations to be more precise). Those who know a lot of calculus just deal with it. Those who don't know any calculus just ignore it.

*R* = ½ρ*CAv*^{2}

A simplified model of drag is one that assumes that drag is directly proportional to speed.. This sometimes is good enough. (Maybe we should call it the" good enough model of drag".) It is especially useful when teaching calculus students how to solve differential equations for the first time. I haven't found it to be all that applicable to real world situations, however. (We'll use *b* as the generic constant of proportionality from now on.)

**R** = − *b***v**

A more general model of drag is one that is agnostic about higher powers (pun intended). This is good attitude to have when you are exploring drag experimentally. Don't assume you know anything about how drag varies with speed, just measure the two quantities and see what values work best for the power *n* and the constant of proportionality *b*.

*R* = − *bv ^{n}*

Possible the mist general model is one that assumes a polynomial relationship. Drag might be related to speed in a way that is partially linear, partially quadratic, partially cubic, and partially described by higher order terms.

*R* = − ∑ *b _{n}v^{n}*

### drag and power

If you want to go fast, you've got to work hard. That should be a statement of the obvious. But why? Well for one thing, it takes energy to get going — kinetic energy. This equation says, if you want to go twice as fast you've go to work four times harder (*K* ∝ *v*^{2}).

*K* = ½*mv*^{2}

While that's certainly true, it isn't of much use to us here on earth. If we lived in the vacuum of space, all we'd ever have to worry about was the energy needed to change our state from one speed to another. Here on earth, the atmosphere has another opinion. Whatever energy we add to a system to get it going, the atmosphere drags it away — all of it eventually. In order for a moving body to stay in motion on the Earth it not only has to get going, it has to actively work to keep going. This undeniable fact of life is why Newton's first law (the law of inertia) wasn't discovered until 1666 (approximately).

To keep an object in motion in the presence of drag (aerodynamic or otherwise) requires an ongoing input of energy. Work must be done over some time. Power must be used. Recall the following chain of reasoning that starts from the definition of power as the rate at which work is done…

P = |
W |
= | F · Δs |
= F · v |

t |
t |

Replace the generic force variable with a generic power equation for drag…

*P* = (*bv ^{n}*)

*v*

Thus in general…

*P* = *bv*^{n + 1}

or more specifically, in the case of pressure drag…

*P* = (½ρ*CAv*^{2}) *v*

*P* = ½ρ*CAv*^{3}

Thus, if drag is proportional to the square of speed, then the power needed to overcome that drag is proportional to the cube of speed (*P* ∝ *v*^{3}). You want to ride your bicycle twice as fast, you'll have to be eight times more powerful. This is why motorcycles are so much faster than bicycles.

Power expended against drag is the biggest impediment to moving freely for both bicycles and motorcycles. Humans can do sustained physical work like cycling at the rate of about a tenth of a horsepower. Motorcycles have engines that are on the order of 100 horsepower. (Sorry for the American units.) That makes a motorcycle about one thousand times more powerful than a human on a bicycle. As a result they can go about ten times faster, since 1,000 = 10^{3}. I've found through personal experience on all day bicycle rides that I typically cover ⅙ the distance that I would if I sat behind the wheel of a car all day.

Yes I realize that cars aren't motorcycles, but what we're really comparing here are wheeled vehicles powered by human muscle with those powered by internal combustion engines. Yes I realize that a 6 to 1 ratio is not exactly the same as 10 to 1, but what I'm doing here is a quick order of magnitude comparison. Your individual results may vary — but not significantly.

### terminal velocity

It's much more than the name of a bad movie. It's something every student of aerodynamic drag should understand.

Imagine yourself as a parachute jumper; or better yet, imagine yourself as a BASE jumper. BASE is an acronym for *b*uilding, *a*ntenna, *s*pan, *e*scarpment. Since none of these platforms is moving horizontally, none of these jumpers has any initial horizontal velocity. Not that it really matters, but this reduces some of the complexity. Step off the platform and draw your free body diagram as you fall.

You start with no initial velocity, there is no aerodynamic drag, and you are effectively in free fall with an acceleration of 9.8 m/s^{2}.

Now it gets complicated. There is an initial acceleration, therefore there is an increase in speed. With an increase in speed comes an increase in drag and a decrease in net force. This decrease in net force reduces acceleration. Speed is still increasing, just not quite as fast as it was initially.

Speed continues to increase, but so too does drag. As drag increases, acceleration decreases. Eventually one can imagine a state when the drag and weight forces are equal. You are in equilibrium. You continue moving, but you cease accelerating. You have reached your terminal velocity. Given the usual posture of skydivers, the type of clothes they normally wear, and the conditions of the air near the surface of the Earth; your typical skydiver has a terminal velocity of 55 m/s (200 km/h or 125 mph). The speed that you have in this state is the one you will always acquire if you are given enough time.

That is until the parachute opens. Opening the chute significantly increases your projected area, which cranks up the aerodynamic drag proportionally. The upward drag force now exceeds the downward pull of gravity. The net force and acceleration are directed upward. Note: this does not mean the skydiver is moving upward. Acceleration does not determine the direction of motion of an object, it determines the direction of the change in motion. When a parachute is just opened, the velocity is down and the acceleration is up. Your speed decreases as a result, which is the whole point behind the parachute.

Speed decreases, so drag decreases. Drag decreases, so the net force decreases. Eventually the net force is zero, you stop accelerating, and you reach a new terminal velocity — one that makes landing more comfortable, something like 6 m/s (22 km/h or 13 mph) or less.

Note that a terminal velocity is not necessarily a maximum value. It's a limit that can be approached from either direction. An object could start off slow and speed up to a terminal velocity that's a maximum (like a skydiver stepping off a BASE) or it could start off fast and slow down to a terminal velocity that's a minimum (like a skydiver who's just opened her parachute). "Terminal" is a fancy way to say "end". A terminal velocity is one that you end with. For falling objects, this occurs when drag equals weight.

R |
= | W |

½ρCAv_{t}^{2} |
= | mg |

v = √ _{t} |
2mg |

ρCA |

Terminal velocity applies to situations besides skydiving. Drive your car with the accelerator in a constant position and you'll eventually reach a terminal velocity. The forward driving force of the tires on the road will eventually equal the backward drag force of the air (and the rolling resistance of the tires, which is discussed somewhere else in this book). Note how I said "eventually". Terminal velocity is a speed things approach but never quite reach. Proof of this statement requires calculus and will be discussed in the practice problems of this section.

Terminal velocity can have any value — including zero. What happens to a ship in the ocean when the propeller stops turning? The forward thrust goes away and all that's left is the backward drag. The ship goes slower and slower and slower until it stops (stops relative to any current, that is). The ship will reach a terminal velocity of zero. For large container ships this may take minutes of time and kilometers of distance, but it will happen eventually. If you don't have the time or the space and you really want to stop a large seagoing vessel, you run the engines in reverse. In this case it's thrust that stops the ship, not drag.

v (m/s)_{t} |
falling object |
---|---|

373 | skydiver, 39 km (Felix Baumgartner, 2012) |

367 | skydiver, 41 km (Alan Eustace, 2014) |

274 | skydiver, 31 km (Joseph Kittinger, 1960) |

146 | skydiver, 04 km (Christian Labhart, 2010) |

55 | skydiver, typical |

45 | bullet |

29 | penny |

25 | cat |

15~40 | hail |

9~13 | raindrop |

6 | skydiver, parachute open |

1~2 | snowflake |

1~2 | ant |